How would one prove that the golden rule is a tautology without a truth table?
$A\,\land\,B\,\equiv\,A\,\equiv\,B\,\equiv\,B\,\lor\,A $
I've been searching a lot and cannot find a demonstration. I've tried to simplify the formula using another inference rules to reach a true value, without success.
Any help or at least a hint would be very appreciated.
$$(A \land B) \equiv A \equiv B \equiv (B \lor A) \Leftrightarrow$$
$$((A \land B \land A) \lor (\neg (A \land B) \land \neg A)) \equiv (((B \land (B \lor A)) \lor (\neg B \land \neg (B \lor A))) \Leftrightarrow $$
$$((A \land B) \lor ((\neg A \lor \neg B) \land \neg A)) \equiv (B \lor (\neg B \land \neg B \land \neg A)) \Leftrightarrow$$
$$((A \land B) \lor \neg A) \equiv (B \lor (\neg B \land \neg A)) \Leftrightarrow $$
$$(B \lor \neg A) \equiv (B \lor \neg A) \Leftrightarrow$$
$$\top$$
I used the following equivalences:
Equivalence
$P \equiv Q \Leftrightarrow (P \land Q) \lor (\neg P \land \neg Q)$
Absorption
$P \land (P \lor Q) \Leftrightarrow P$
$P \lor (P \land Q) \Leftrightarrow P$
Reduction
$P \land (\neg P \lor Q) \Leftrightarrow P \land Q$
$P \lor (\neg P \land Q) \Leftrightarrow P \lor Q$
Biconditional Tautology
$P \equiv P \Leftrightarrow \top$